给定\(n\)个点，\(m\)条边的带权无向图

选出一些边，使得\(4\)对点之间可达，询问权值最小为多少

\(n \leqslant 30, m \leqslant 1000\)

首先看数据范围，\(4\)对点，也就是\(8\)个点，很小

上斯坦纳树（局部最小生成树）

然而好像题目并不是斯坦纳树，可能是一些树拼到一起

那么就再做一个状压\(dp\)即可

复杂度\(O(3^8 * n + 2^8 * nm + 2^{12} * n)\)

#include <map>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;const int sid = 55;
const int eid = (1 << 9) - 1;
inline void cmin(int &a, int b) { if(a > b) a = b; }int n, m, nc;
int U[sid], V[sid], cl[sid], ip[sid];
int f[sid][eid], E[sid][sid];
string s[55], sa, sb;
map <string, int> id;int inq[sid];
queue <int> q;
void spfa(int S) {memset(inq, 0, sizeof(inq));for(int i = 1; i <= n; i ++) q.push(i);while(!q.empty()) {int id = q.front(); q.pop(); inq[id] = 0;for(int j = 1; j <= n; j ++)if(f[j][S] > f[id][S] + E[id][j]) {f[j][S] = f[id][S] + E[id][j];if(!inq[j]) q.push(j), inq[j] = 1;}}
}void wish1() {memset(f, 56, sizeof(f));for(int i = 1; i <= n; i ++)if(cl[i]) {nc ++;ip[i] = nc;f[i][(1 << nc - 1)] = 0;}for(int S = 0; S <= (1 << nc) - 1; S ++) {for(int i = 1; i <= n; i ++)for(int T = S; T; T = (T - 1) & S)cmin(f[i][S], f[i][T] + f[i][S ^ T]);spfa(S);}
}int g[eid];
void wish2() {memset(g, 56, sizeof(g)); g[0] = 0;for(int s = 0; s <= (1 << 4) - 1; s ++)for(int i = 1; i <= n; i ++)for(int S = 0; S <= (1 << nc) - 1; S ++) {int T = 0;for(int k = 1; k <= 4; k ++)if((S & (1 << ip[U[k]] - 1)) && (S & (1 << ip[V[k]] - 1)))T |= (1 << k - 1);cmin(g[s | T], g[s] + f[i][S]);}printf("%d\n", g[(1 << 4) - 1]);
}int main() {freopen("bzoj1402.in", "r", stdin);freopen("bzoj1402.out", "w", stdout);cin >> n >> m;for(int i = 1; i <= n; i ++) {cin >> s[i];id[s[i]] = i;}memset(E, 56, sizeof(E));for(int i = 1; i <= m; i ++) {int u, v, w;cin >> sa >> sb >> w;u = id[sa]; v = id[sb];cmin(E[u][v], w); E[v][u] = E[u][v];}for(int i = 1; i <= 4; i ++) {int u, v;cin >> sa >> sb;u = id[sa]; v = id[sb];U[i] = u; V[i] = v;cl[u] = cl[v] = 1;}wish1(); wish2();return 0;
}